16
$\begingroup$

Assume that $X_1,\cdots,X_n$ are independent r.v., not necessarily iid,

Let $S_n=X_1+\cdots+X_n$, suppose that $S_n$ converges in probability, how can we show that $S_n$ converges a.s.?

$\endgroup$
2
  • $\begingroup$ I am not sure whether the following would help. If $A$ is the set of $\omega$ for which $S_n(\omega)$ converges, by Kolmogorov 0-1 law, $P(A)=0$ or $1$. $\endgroup$
    – Ashok
    Dec 10 '11 at 7:49
  • 7
    $\begingroup$ See Theorem 5.3.4 in Kai Lai Chung, A course in probability theory, 3rd ed., Academic Press, 2001. $\endgroup$
    – Did
    Oct 31 '12 at 20:53
14
$\begingroup$

We can use Ottaviani's inequality: if we put $M_k:=\max_{1\leqslant i\leqslant k}|S_i|$ and $S_{k,n}:=\sum_{i=k+1}^nX_i$, then for all $\varepsilon >0$ we have $$\min_{1\leqslant k\leqslant n}P(|S_{k,n}|\leqslant\varepsilon)P(|M_n|>2\varepsilon)\leqslant P(|S_n|>\varepsilon).$$

Put $A_m:=\sup_{k\in\mathbb N^*}|S_{m+k}-S_m|$ and $A:=\inf_{m\in\mathbb N^*}A_m$. We have $$\{\{S_n\}\mbox{ doesn't converge}\}=\{A\neq 0\}\subset\bigcup_{\varepsilon\in\mathbb Q^+}\bigcap_{m\in\mathbb N^*}\{A_m>\varepsilon\}$$ and $$\{A_m>\varepsilon\}=\bigcup_{r\in\mathbb N^*}\left\{\sup_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>\varepsilon\right\}.$$ Ottaviani's inequality gives $$\min_{1\leqslant k\leqslant r}P(|S_{m+r}-S_{m+k}|\leqslant\varepsilon)P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_m|>2\varepsilon\right)\leqslant P(|S_{r+m}-S_m|>\varepsilon).$$ We fix $\delta>0$; we can choose $N=N(\varepsilon,\delta)$ such that for $0\leqslant k\leqslant r$ and $m\geqslant N$, $P(|S_{m+r}-S_{m+k}|>\varepsilon)\leqslant\delta$. We get $$P\left(\max_{1\leqslant k\leqslant r}|S_{m+k}-S_{m+r}|>\varepsilon\right)\leqslant\frac{\delta}{1-\delta},$$ therefore $P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta}$. We have $$0\leqslant P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)\leqslant P(A_m>\varepsilon)\leqslant\frac{\delta}{1-\delta},$$ so for all $\varepsilon\in\mathbb Q^+$, we have $P\left(\bigcap_{p\in\mathbb N^*}(A_p>\varepsilon)\right)=0$, hence $P(A\neq 0)=0$ and we have the almost sure convergence.

$\endgroup$
1
  • $\begingroup$ great.....I tried without Ottavani.....Ottavani is necessary else we get a bound of N$\delta$ and cant control the probabilities $\endgroup$
    – user24367
    Feb 17 '13 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.